College Algebra, 2013a

3.4 Complex Zeros and the Fundamental Theorem of Algebra 293
2+2i 1 0 0 0 64
↓ 2+2i 8i −16 + 16i −64
2 − 2i 1 2+2i 8i −16 + 16i 0
↓ 2 − 2i 8 − 8i 16 − 16i
1 4 8 0
Our quotient polynomial is x 2 +4x+8. Using the quadratic formula, we obtain the remaining
zeros −2+2i and −2 − 2i.
3. Using Theorem 3.14, wegetf(x) =(x − [2 − 2i])(x − [2 + 2i])(x − [−2+2i])(x − [−2 − 2i]).
4. We multiply the linear factors of f(x) which correspond to complex conjugate pairs. We find
(x − [2 − 2i])(x − [2 + 2i]) = x 2 − 4x +8,and(x − [−2+2i])(x − [−2 − 2i]) = x 2 +4x +8.
Our final answer is f(x) = ( x 2 − 4x +8 )( x 2 +4x +8 ) .
Our last example turns the tables and asks us to manufacture a polynomial with certain properties
of its graph and zeros.
Example 3.4.4. Find a polynomial p of lowest degree that has integer coefficients and satisfies all
of the following criteria:
ˆ the graph of y = p(x) touches (but doesn’t cross) the x-axis at ( 1
3 , 0)
ˆ x =3i is a zero of p.
ˆ as x →−∞, p(x) →−∞
ˆ as x →∞, p(x) →−∞
Solution. To solve this problem, we will need a good understanding of the relationship between
the x-intercepts of the graph of a function and the zeros of a function, the Factor Theorem, the
role of multiplicity, complex conjugates, the Complex Factorization Theorem, and end behavior of
polynomial functions. (In short, you’ll need most of the major concepts of this chapter.) Since the
graph of p touches the x-axis at ( 1
3 , 0) ,weknowx = 1 3
is a zero of even multiplicity. Since we
are after a polynomial of lowest degree, we need x = 1 3
to have multiplicity exactly 2. The Factor
Theorem now tells us ( x − 1 2
3)
is a factor of p(x). Since x =3i is a zero and our final answer is to
have integer (real) coefficients, x = −3i is also a zero. The Factor Theorem kicks in again to give us
(x−3i) and(x+3i) as factors of p(x). We are given no further information about zeros or intercepts
so we conclude, by the Complex Factorization Theorem that p(x) =a ( x − 1 2
3)
(x − 3i)(x +3i) for
some real number a. Expanding this, we get p(x) =ax 4 − 2a 3 x3 + 82a
9 x2 −6ax+a. In order to obtain
integer coefficients, we know a must be an integer multiple of 9. Our last concern is end behavior.
Since the leading term of p(x) isax 4 , we need a